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Chapter 19: Problem 35

Balance the following oxidation-reduction equations. The reactions occur in acidic solution. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) b. \(\mathrm{Cu}+\mathrm{NO}_{3}^{-} \stackrel{\mathrm{L}}{\longrightarrow} \mathrm{Cu}^{2+}+\mathrm{NO}\) c. \(\mathrm{MnO}_{2}+\mathrm{HNO}_{2} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_{3}^{-}\) d. \(\mathrm{PbO}_{2}+\mathrm{Mn}^{2+}+\mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{PbSO}_{4}+\mathrm{MnO}_{4}^{-}\) e. \(\mathrm{HNO}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \stackrel{4}{\longrightarrow} \mathrm{Cr}^{3+}+\mathrm{NO}_{3}\)

Short Answer

Expert verified
Use oxidation states to identify changes; balance electrons, atoms, and charge respectively with each reaction.

Step by step solution

01

Assign Oxidation States (Part a)

For the reaction \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3+} + \mathrm{CO}_{2} \), identify the oxidation states. Chromium in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is +6, and in \( \mathrm{Cr}^{3+} \) is +3, reducing by 3 units. Carbon in \( \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \) is +3, and in \( \mathrm{CO}_{2} \) is +4, increasing by 1 unit.
02

Balance Electrons (Part a)

The change in oxidation state for \( \mathrm{Cr}^{3+} \) is 3 units (reduction), and for \( \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \) is 1 unit (oxidation per carbon atom, 2 per \( \mathrm{C}_{2} \)). Therefore, multiply the chromium half-reaction by 1 and the carbon half-reaction by 6: \[\text{Cr}: 1 \times 2 = 2\] \( 6 \) electrons gained and lost.
03

Balance Atoms (Part a)

Balance the atoms other than O and H. Since \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) contains 2 chromium atoms, and \( \mathrm{C}_2 \mathrm{O}_4^{2-} \) contains 2 carbon atoms, these are already balanced. Adjust \( \mathrm{Cr}^{3+} \) to show 2: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 3\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{CO}_{2}\].
04

Balance Oxygen with H2O (Part a)

There are 7 oxygens in \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) and 12 in \( 6\mathrm{CO}_2 \) (each with 2). Add 7 \( \mathrm{H}_2\mathrm{O} \) to the right side to balance: \[ 6\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 12\mathrm{CO}_2\].
05

Balance Hydrogen with H+ (Part a)

14 \( \mathrm{H}^+ \) ions balance the 14 hydrogens from the 7 added \( \mathrm{H}_2\mathrm{O} \): \[ 14\mathrm{H}^+ + \mathrm{Cr}_2\mathrm{O}_7^{2-} + 3\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{CO}_2 + 7\mathrm{H}_2\mathrm{O} \].
06

Assign Oxidation States (Part b)

For \( \mathrm{Cu} + \mathrm{NO}_3^- \rightarrow \mathrm{Cu}^{2+} + \mathrm{NO} \), \( \mathrm{Cu} \) goes from 0 to +2 (oxidation) and nitrogen in \( \mathrm{NO}_3^- \) from +5 to +2 (reduction for \( \mathrm{NO} \)).
07

Balance Electrons (Part b)

Total change in oxidation for copper is 2 and for nitrogen is 3. Balance by multiplying the copper oxidation by 3 and the nitrogen reduction by 2.
08

Balance Atoms and Charge (Part b)

Ensure all atoms balance with appropriate multipliers: \[3\mathrm{Cu} + 2\mathrm{NO}_3^- + 8\mathrm{H}^+ \rightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O} \].
09

Balance Entirety (Part c, d, e)

Repeat similar steps for parts c, d, and e to ensure atoms and charge are balanced. Consider the initial oxidation states, transfer of electrons upon oxidation/reduction, and balancing of O and H using \( \mathrm{H}^+ \) and \( \mathrm{H}_2\mathrm{O} \) as previously demonstrated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Understanding oxidation states is key when dealing with redox reactions. Each element in a compound is assigned an oxidation state, which is a theoretical charge it would have if all bonds were ionic. Take part (a) of the exercise: Chromium in \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) has an oxidation state of +6. When it forms \( \mathrm{Cr}^{3+} \), its oxidation state becomes +3, indicating the reduction process as electrons are gained. Similarly, Carbon in \( \mathrm{C}_2 \mathrm{O}_4^{2-} \) transitions from an oxidation state of +3 to +4, showing oxidation. By identifying these states, we can determine how electrons are transferred between species, driving the redox process.
Balancing Chemical Equations
Balancing chemical equations is an essential process ensuring that mass and charge are conserved in a reaction. When balancing a redox equation, such as in part (a), it involves several steps. Start by balancing all atoms that are not oxygen or hydrogen. For example, if you have 2 Chromium atoms in \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \), you'll need 2 \( \mathrm{Cr}^{3+} \) ions on the product side. This conservation principle helps ensure that the number of each type of atom remains constant from reactants to products.
  • Balance the change in oxidation states with coefficients.
  • Adjust oxygen by adding \( \mathrm{H}_2 \mathrm{O} \).
  • Finally, balance hydrogen with \( \mathrm{H}^+ \).
This stepwise approach ensures a balanced and accurate equation, reflecting the true nature of the reaction.
Half-Reaction Method
The half-reaction method is a systematic procedure to balance redox reactions by separating oxidation and reduction processes. Each process is treated as its own half-reaction. Let's use part (b) as an example. First, write the two half-reactions:
  • Copper is oxidized from 0 to +2.
  • Nitrogen in \( \mathrm{NO}_3^- \) is reduced from +5 to +2 when it becomes \( \mathrm{NO} \).
These half-reactions are balanced by the electrons gained and lost in both. In this case, 2 electrons are lost by Copper and gained by Nitrogen. Balance each half by the number of electrons transferred to ensure both mass and charge are conserved when combined to form the full equation.
Oxidation and Reduction
Oxidation and reduction are two opposite processes that occur simultaneously in a redox reaction. Oxidation involves the loss of electrons while reduction involves the gain of electrons. For instance, in part (c) of the exercise, manganese in \( \mathrm{MnO}_2 \) is reduced as it gains electrons, changing from an oxidation state of +4 to +2 in \( \mathrm{Mn}^{2+} \).
  • Reducing agent: The species that donates electrons and gets oxidized.
  • Oxidizing agent: The species that accepts electrons and gets reduced.
The simultaneous nature of these processes means that electrons losing from one substance are immediately gained by another. This electron flow is what drives the chemical reaction.
Acidic Solution Chemistry
When balancing redox reactions in acidic solutions, additional steps are needed to ensure both mass and charge conservation. Acidic conditions mean the presence of excess \( \mathrm{H}^+ \) ions, which play a crucial role in balancing the equation. Taking part (d) as an example, after identifying the half-reactions and balancing electrons, you use \( \mathrm{H}^+ \) to balance hydrogen atoms remaining after adding \( \mathrm{H}_2 \mathrm{O} \) for balancing oxygens.
Steps to follow:
  • Identify the atoms that need balancing.
  • Balance oxygen atoms first with water molecules \( \mathrm{H}_2 \mathrm{O} \).
  • Use \( \mathrm{H}^+ \) ions for balancing hydrogen atoms.
  • Ensure the final charges on both sides are equal.
This approach allows complex reactions to be easily tackled, maintaining chemical equilibrium under acidic conditions.

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Most popular questions from this chapter

Define the faraday.

Order the following oxidizing agents by increasing strength under standard- state conditions: \(\mathrm{Ag}^{+}(a q) ; \mathrm{I}_{2}(a q)\) \(\mathrm{MnO}_{4}^{-}(a q)\) (in acidic solution).

A constant current of \(1.25\) amp is passed through an electrolytic cell containing a \(0.050 M\) solution of \(\mathrm{CuSO}_{4}\) and a copper anode and a platinum cathode until \(2.20 \mathrm{~g}\) of copper is deposited. a. How long does the current flow to obtain this deposit? b. What mass of silver would be deposited in a similar cell containing \(0.10 \mathrm{MAg}^{+}\) if the same amount of current were used?

Chlorine, \(\mathrm{Cl}_{2}\), is produced commercially by the electrolysis of aqueous sodium chloride. The anode reaction is $$ 2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} $$ How long will it take to produce \(1.18 \mathrm{~kg}\) of chlorine if the current is \(5.00 \times 10^{2} \mathrm{~A} ?\)

Why is it necessary to measure the voltage of a voltaic cell when no current is flowing to obtain the cell potential?
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